Partial sum formula of following series $$\sum_{m \ge 1} \frac{(xy)^m}{(2m)!(1-y^m)}, \quad\text{where }x,y \in \mathbb N$$
I have, to start, J.Jacquelin's answer. 
 A: This is a partial answer giving a double path-integral over the function defined in (1).
Let us write for simplicity
$$H(x,y) := \sum_{m=1}^\infty \frac{(xy)^m}{(2m)!(1-y^m)}.$$
This sum consits of all diagonal terms of the double series
$$G(x,y):=\sum_{m=1}^\infty \frac{x^m}{(2m)!} \sum_{n=1}^\infty \frac{y^n}{1-y^n}.$$
For $x \in \mathbb{C}_{-} := \mathbb{C} \setminus \{ x \in \mathbb{R} : x \le 0 \}$ we have
$$\sum_{m=1}^\infty \frac{x^m}{(2m)!} = \cosh(\sqrt{x}) -1$$
where we take the the main branch for the square root. We also have for $y$ with $ 0< \mathrm{Re}(y) < 1$
$$ \sum_{n=1}^\infty \frac{y^n}{1-y^n} = \frac{\psi_y(1) + \log(1-y)}{\log(y)},$$
where $\psi_q$ denotes the q-Polygamma function. Thus
$$\tag{1} G(x,y) = \{ \cosh(\sqrt{x}) -1 \} \frac{\psi_y(1) + \log(1-y)}{\log(y)}.$$
Now, we can apply the Cauchy integral formula to get
\begin{align}
H(x,y) &= \frac{1}{2\pi i} \sum_{n=1}^\infty \int_{ \partial B_\varepsilon(x)} \int_{\partial B_\varepsilon(y)} \frac{G(z,w)}{(z-x)^n (w-y)^n} \, dw dz \\
&= \frac{1}{2\pi i} \int_{\partial B_\varepsilon(x)} \int_{\partial B_\varepsilon(y)} \frac{G(z,w)}{(z- x)(w-y)-1} \, dw dz.
\end{align}
At this point, one may use the residue theorem.
